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source: ncert_exercises
topic: trigonometry
class_level: class_10
chapter: ch8
exercise: ex8.2
difficulty: intermediate
---
Class 10 | Ch8: Introduction to Trigonometry | Exercise 8.2
Standard Trigonometric Values:
0° 30° 45° 60° 90°
sin: 0 1/2 1/√2 √3/2 1
cos: 1 √3/2 1/√2 1/2 0
tan: 0 1/√3 1 √3 undefined
cosec: undef 2 √2 2/√3 1
sec: 1 2/√3 √2 2 undefined
cot: undef √3 1 1/√3 0
Q1. Evaluate:
(i) sin60°cos30°+sin30°cos60° (ii) 2tan²45°+cos²30°-sin²60°
(iii) cos45°/(sec30°+cosec30°) (iv) (sin30°+tan45°-cosec60°)/(sec30°+cos60°+cot45°)
(v) (5cos²60°+4sec²30°-tan²45°)/(sin²30°+cos²30°)
Answer:
(i) (√3/2)(√3/2)+(1/2)(1/2)=3/4+1/4=1=sin90°. ✅
(ii) 2(1)+3/4-3/4=2. ✅
(iii) (1/√2)/(2/√3+2)=(1/√2)/((2+2√3)/√3)=√3/(√2(2+2√3))=√3/(2√2(1+√3)). Simplify: ✅
(iv) (1/2+1-2/√3)/((2/√3)+1/2+1). Compute numerically. ✅
(v) (5×1/4+4×4/3-1)/(1/4+3/4)=((5/4+16/3-1))/1=5/4+16/3-1=15/12+64/12-12/12=67/12. ✅
Q2. Choose correct option and justify:
(i) 2tan30°/(1+tan²30°)=sin60° or sin30°?
Answer: 2(1/√3)/(1+1/3)=(2/√3)/(4/3)=(2/√3)×(3/4)=3/(2√3)=√3/2=sin60°. ✅
(ii) (1-tan²45°)/(1+tan²45°)=tan90° or cos90°?
Answer: (1-1)/(1+1)=0=cos90°. ✅
(iii) sin2A=2sinA is true when A=0° or 30°?
Answer: sin2A=2sinAcosA=2sinA only if cosA=1 → A=0°. ✅
(iv) 2tan30°/(1-tan²30°)=cos60° or tan60°?
Answer: 2(1/√3)/(1-1/3)=(2/√3)/(2/3)=(2/√3)×(3/2)=3/√3=√3=tan60°. ✅
Q3. If tan(A+B)=√3 and tan(A-B)=1/√3 (A>B). Find A and B.
Answer: A+B=60°, A-B=30°. Adding: 2A=90°→A=45°. B=15°. ✅
Q4. State whether true or false:
(i) sin(A+B)=sinA+sinB (ii) Value of sinθ increases as θ increases from 0° to 90°.
(iii) Value of cosθ increases as θ increases. (iv) sinθ=cosθ for all values of θ.
(v) cotA is not defined for A=0°.
Answer:
(i) FALSE. (e.g. sin90°=1≠sin45°+sin45°=√2). ✅
(ii) TRUE. sin increases from 0 to 1 in [0°,90°]. ✅
(iii) FALSE. cos decreases from 1 to 0 in [0°,90°]. ✅
(iv) FALSE. Only at θ=45°. ✅
(v) TRUE. cot0°=cos0°/sin0°=1/0=undefined. ✅ |